To solve this problem, we first note the key details:

Problem Breakdown:

1. A standard 6-sided die contains numbers $1, 2, 3, 4, 5, 6$.
2. The die is rolled twice, so the possible sums range from $2$ (1+1) to $12$ (6+6).
3. We need to find the maximum number of ways to get a sum that has exactly 2 factors, meaning the sum must be a prime number.

Step 1: Identify the prime numbers in the range

The prime numbers between $2$ and $12$ are: $2, 3, 5, 7, 11$

Step 2: Determine the pairs of dice rolls for each prime sum

Now, for each prime sum, find the combinations of rolls that result in that sum:

Sum = 2

• $(1, 1)$ → 1 way

Sum = 3

• $(1, 2), (2, 1)$ → 2 ways

Sum = 5

• $(1, 4), (2, 3), (3, 2), (4, 1)$ → 4 ways

Sum = 7

• $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ → 6 ways

Sum = 11

• $(5, 6), (6, 5)$ → 2 ways

Step 3: Calculate the maximum

The sum $7$ has the maximum number of combinations (6 ways).

Final Answer:

The maximum number of ways for the sum of the two recorded numbers to be a number with exactly 2 factors is 6.

Would you like a detailed explanation of why these sums correspond to specific dice rolls, or help with related questions? Here are five questions for further exploration:

1. What are the total number of outcomes when rolling two dice?
2. How can the total probability of rolling a prime number as the sum be calculated?
3. What is the probability of rolling a sum of $7$?
4. Can this process be extended to dice with more faces (e.g., an 8-sided die)?
5. How do probabilities change when restricting outcomes to odd sums only?

Tip: Prime numbers are key in many probability problems because they often limit the number of viable combinations—an important tool in problem-solving